Question

How do I solve this three variable equation

Answer 1

A) 2.x - 6.y + z = 30
b) -x  + 3.y +2.z = - 20
c) -6.x -5.y +6.z = 7

1st, let's take a) and b) alone and let's eliminate x

a) 2.x - 6.y + z = 30
b) -x  + 3.y +2.z = - 20

2nd) multiply b) with 2 and once done add a) with the result of b):
b') times 2 → -2.x+6.y+4.z = - 40. Now add it to a):

a) 2.x - 6.y + z = 30
b') -2.x+6.y+4.z = - 40
+  -----------------    -----
     0.x + 0.y + 5.z = -10    → z = -2

Now replace z with its value (-2) in :

b) -x  + 3.y +2.z = - 20  → - x + 3.y -4 = -20 → - x +3.y = -16
c) -6.x -5.y +6.z = 7 → -6.x - 5.y -12 = 7 → - 6.x -5.y = 19

Consider the result of:
- x +3.y = -16  and
- 6.x -5.y = 19
Now let's eliminate x from both equations.
To that end multiply the 1st one by (-6)
6x - 18y = 96  and add it to - 6.x -5.y = 19

6x - 18y = 96
-6.x -5.y = 19
+---------    ----
0.x - 23.y   115  → 23y = -115 and y = -115/23 = - 5→→ y = -5

Now that you have the value of z= - 2 and y = -5, plugin in any of the equation and you will find x = 1. So the final result is:
 x = 1 ; y = - 5  ; and z = - 2

Answer 2

2x - 6y + z = 30      (1st)
-x + 3y + 2z = -20 (2nd)
-6x - 5y + 6z = 7   (3rd)

from (2nd)  you have:
-x + 3y + 2z = -20 
 x = 3y + 2z + 20

substitute x = 3y + 2z + 20 into  (1st) 2x - 6y +z = 30

2x - 6y + z = 30
2(3y + 2z + 20) - 6y +z = 30
6y + 4z + 40 - 6y + z = 30
5z + 40 = 30
5z = -10
  z = -2

plug in z = - 2 into (2nd) and  (3rd)

-x + 3y + 2(-2) = -20 (2nd)
-x + 3y = -16 so x = 3y + 16

-6x - 5y + 6z = 7   (3rd)
-6x -5y +6(-2) = 7
-6x - 5y = 19

substitute x = 3y + 16 into -6x - 5y = 19

-6x - 5y = 19
-6(3y + 16) - 5y = 19
-18y - 96 - 5y = 19
-23y = 115
     y = -5

x = 3y + 16
x = 3(-5) + 16
x = -15 + 16
x = 1

answer
x = 1, y = -5 and z = -2